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Use the Zero Product Property.
Use the Zero Product Property.
x=4 or x=- 10
x=- 8 or x=3/2
We want to solve the given equation for x. To do this, we can start by using factoring. Then, we will use the Zero Product Property.
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
x^2+6x- 40=0
In this case, we have - 40. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign (one positive and one negative.)
| Factor Constants | Product of Constants |
|---|---|
| - 1 and 40 | - 40 |
| 1 and -40 | - 40 |
| - 2 and 20 | - 40 |
| 2 and -20 | - 40 |
| - 4 and 10 | - 40 |
| 4 and -10 | - 40 |
| - 5 and 8 | - 40 |
| 5 and -8 | - 40 |
Next, let's consider the coefficient of the linear term. x^2+6x- 40=0 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 6.
| Factors | Sum of Factors |
|---|---|
| - 1 and 40 | 39 |
| 1 and -40 | - 39 |
| - 2 and 20 | 18 |
| 2 and -20 | - 18 |
| - 4 and 10 | 6 |
| 4 and -10 | - 6 |
| - 5 and 8 | 3 |
| 5 and -8 | - 3 |
We found the factors whose product is - 40 and whose sum is 6. x^2+6x- 40=0 ⇕ (x-4)(x+10)=0
Since the equation is already written in factored form, we can now use the Zero Product Property.
Use the Zero Product Property
(I): LHS+4=RHS+4
(II): LHS-10=RHS-10
We found that x=4 or x=- 10.
We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=4.
x= 4
Calculate power
Multiply
Add and subtract terms
Substituting and simplifying created a true statement, so we know that x=4 is a solution of the equation. Let's move on to x=- 10.
x= - 10
(- a)^2 = a^2
a(- b)=- a * b
Add and subtract terms
Again, we created a true statement. x=- 10 is indeed a solution of the equation.
We want to solve the given equation for x. To do this, we can start by using factoring. Then, we will use the Zero Product Property.
Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b.
2x^2+13x-24=0
⇕
2x^2+13x+(- 24)=0
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &48 &-1 + 48 &47 - 2 &24 &-2 + 24 &22 - 3 & 16 & - 3 + 16 &13 - 4 &12 &-4 + 12 &8 - 6 &8 &-6 + 8 &2
Finally, we will factor the last expression obtained.
Factor out x
Factor out 8
Factor out (2x-3)
Since the equation is already written in factored form, we can now use the Zero Product Property.
Use the Zero Product Property
(I): LHS-8=RHS-8
We found that x=- 8 or x= 32.
We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=- 8.
x= - 8
(- a)^2 = a^2
a(- b)=- a * b
Multiply
Subtract terms
Substituting and simplifying created a true statement, so we know that x=- 8 is a solution of the equation. Let's move on to x= 32.
x= 3/2
(a/b)^m=a^m/b^m
Calculate power
a*b/c= a* b/c
a/b=a * 2/b * 2
Add fractions
Calculate quotient
Subtract term
Again, we created a true statement. x= 32 is indeed a solution of the equation.